A closed orgain pipe has length $'l’$. The air in it is vibrating in $3^{rd}$ overtone with maximum displacement amplitude $'a’$. The displacement amplitude at distance $l / 7$ from closed end of the pipe is:
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closed organ pipe is in third overtone so total length will be equal to $\frac{\lambda \times 7}{4} .$ so, $\frac{\lambda \times 7}{4}=L$ $\frac{L}{7}=\frac{\lambda}{4}$
amplitude at $\frac{\lambda}{4}$ from the closed end is maximum
so amplitude at $\frac{L}{7}$ is $a.$
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